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130+(x^2+5x)=180
We move all terms to the left:
130+(x^2+5x)-(180)=0
We add all the numbers together, and all the variables
(x^2+5x)-50=0
We get rid of parentheses
x^2+5x-50=0
a = 1; b = 5; c = -50;
Δ = b2-4ac
Δ = 52-4·1·(-50)
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{225}=15$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-15}{2*1}=\frac{-20}{2} =-10 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+15}{2*1}=\frac{10}{2} =5 $
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